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Chapter 12: Problem 53

A cubic centimeter of the air you breathe contains about \(10^{19}\) molecules. A cubic centimeter of a comet's tail may typically contain 200 molecules. Calculate the cubic volume of comet tail material that would hold \(10^{19}\) molecules.

Short Answer

Expert verified
The cubic volume is \(5 \times 10^{16}\) cm^3.

Step by step solution

01

- Understand the given data

There are approximately \(10^{19}\) molecules in one cubic centimeter of air and 200 molecules in one cubic centimeter of a comet's tail.
02

- Define the variables

Let \(V\) be the volume in cubic centimeters of the comet's tail that would contain \(10^{19}\) molecules.
03

- Set up the equation

Since each cubic centimeter of the comet's tail contains 200 molecules, the volume required to hold \(10^{19}\) molecules can be represented by: \[ V \times 200 = 10^{19} \]
04

- Solve for V

To find \(V\), divide both sides of the equation by 200:\[ V = \frac{10^{19}}{200} \]
05

- Simplify the expression

Simplify the division: \[ 10^{19} \div 200 = 5 \times 10^{16} \]
06

- Verify the result

Check the calculation by multiplying \(5 \times 10^{16}\) by 200 to ensure it equals \(10^{19}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Density
Molecular density is the number of molecules present in a given volume of a substance. In our exercise, we compared the molecular density of air and a comet's tail.
A cubic centimeter of air contains about \(10^{19}\) molecules. This is a very high density compared to the comet's tail, which only has 200 molecules per cubic centimeter.
Understanding molecular density helps in calculating how the same number of molecules fits into different volumes, depending on the substance's density.
Volume Calculation
To find the volume required for a certain number of molecules in a different environment, you need to understand volume calculation.
In the original exercise, we aimed to find the volume of the comet's tail that can contain \(10^{19}\) molecules. Given that one cubic centimeter of the comet's tail holds 200 molecules, we set up an equation: \[ V \times 200 = 10^{19} \] Dividing both sides by 200, we get: \[ V = \frac{10^{19}}{200} \] Solving this, we find that \[ V = 5 \times 10^{16} \] Therefore, it takes \(5 \times 10^{16}\) cubic centimeters of comet's tail material to hold \(10^{19}\) molecules.
Scientific Notation
Scientific notation makes it easier to handle very large or very small numbers. It's especially useful in scientific calculations.
In our problem, we deal with numbers like \(10^{19}\) and 200. Using scientific notation, \(10^{19}\) stays the same but when we divide it by 200, we can express the result as: \[ 10^{19} \div 200 = 5 \times 10^{16} \] This simplified form allows for clearer and more manageable calculations.
Problem-Solving
Breaking a problem into steps simplifies complex calculations. The original exercise was solved in several steps:
- Understanding the given data
- Defining the variables
- Setting up an equation
- Solving and simplifying the equation
Each step helps guide the problem-solving process, ensuring that you tackle each part methodically. In our case, solving for volume by dividing the number of molecules by their density in the comet's tail (200 molecules per cm³) led us to a straightforward answer: \[ V = 5 \times 10^{16} \] Always verify your results to ensure your calculations are correct.

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Most popular questions from this chapter

T/F: Meteorites are often found on the ice in Antarctica.

Describe differences between the Kuiper Belt and the Oort Cloud as sources of comets. What is the ultimate fate of a comet from each of these reservoirs?

As a comet leaves the inner Solar System, the ion tail points a. back along the orbit. b. forward along the orbit. c. toward the Sun. d. away from the Sun.

Most meteors that you see began as meteoroids the size of a. tiny dust grains. b. a pencil. c. a car. d. a house.

One recent estimate concludes that nearly 800 meteorites with mass greater than 100 grams (massive enough to cause personal injury strike the surface of Earth each day. Assuming you present a target of 0.25 square meter \(\left(\mathrm{m}^{2}\right)\) to a falling meteorite, what is the probability that you will be struck by a meteorite during your 100 -year lifetime? (Note that the surface area of Earth is approximately \(5 \times 10^{14} \mathrm{m}^{2} .\) )
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